Behavior of wire subjected by two equal and opposite pulling forces?

In summary, we discussed the calculation of work done by a wire under tension, the difference between forces and extensions in Hooke's law, the proper mathematical representation of wire extension, the change in energy stored in the wire, when to consider changes in cross-sectional area, and the necessity of the approximation e^2 ≈ 0 when calculating stored energy.
  • #1
johnconnor
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Imagine a wire with uniform circular cross-sectional area and length l. It is being pulled at both ends with a horizontal force that is equal and opposite to each other. The wire obeys Hooke's law up to breaking.

(1) How do we calculate the work done by the wire if both sides of the wire extend by 2e (i.e. an extension of "e" at each end of the wire)? (2) How do I explain the difference between the cancellation of the vector components of each force and the fact that the wire still extends? (3) What is the most proper way of writing down the mathematical working of the extension of the wire? On a side note, (4) How does the energy stored in the wire change, then?

(5) Also, in what cases are we supposed to assume that the cross-sectional area remains approximately the same and when do we not? (6) Is the approximation e^2 ≈ 0 necessary in this case when I'm calculating the energy stored in the wire?

Thank you for your help.
 
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  • #2
johnconnor said:
Imagine a wire with uniform circular cross-sectional area and length l. It is being pulled at both ends with a horizontal force that is equal and opposite to each other. The wire obeys Hooke's law up to breaking.

(1) How do we calculate the work done by the wire if both sides of the wire extend by 2e (i.e. an extension of "e" at each end of the wire)?
You can use [itex] W = \int F.ds[/itex], where F is the tension force and ds is the infinitesimal extension. Since the tension force varies from 0 to T during the extension of 2e, the integral works out to (T/2)(2e) = Te. Or if you don't like integrals, the wire is a spring with force constant k = T/2e (per Hookes law F=kx), so you can use energy methods [itex]W = \Delta PE = 1/2kx^2 = (T/(4e))(4e^2) = Te [/itex], same result.
2) How do I explain the difference between the cancellation of the vector components of each force and the fact that the wire still extends?
The forces must be equal and opposite for equilibrium. Internally, the applied force T, at one end, must be equal to the internal force, T, in the rope, for equilibrium
(3) What is the most proper way of writing down the mathematical working of the extension of the wire?
extension = FL/AE, where F = the internal tension force and A is the wire cross section and E is the elasticity modulus (note that k=AE/L) , if that is what you mean
On a side note, (4) How does the energy stored in the wire change, then?
its strain energy increases due to the increase of its PE = 1/2kx^2
(5) Also, in what cases are we supposed to assume that the cross-sectional area remains approximately the same and when do we not?
When the wire obeys hookes law in its elastic range, ignore any cross section change, which only becomes significant in the plastic range beyond yield when the wire will 'neck' down
(6) Is the approximation e^2 ≈ 0 necessary in this case when I'm calculating the energy stored in the wire?
as long as e is such that the wire still obeys Hooke's law, then the stored energy is valid. Since e is on the order of .001 L, I guess you are right that e^2 must be small
Thank you for your help.
'welcome.
 

1. What is the behavior of a wire subjected to two equal and opposite pulling forces?

The behavior of a wire subjected to two equal and opposite pulling forces is known as tension. This means that the wire will experience a stretching force in both directions, causing it to elongate and become thinner.

2. How does the type of wire affect its behavior under two equal and opposite pulling forces?

The type of wire can greatly affect its behavior under two equal and opposite pulling forces. Different materials have different elastic properties, meaning some wires will be more resistant to stretching than others. Additionally, the thickness and length of the wire can also impact its behavior.

3. Can a wire break under two equal and opposite pulling forces?

Yes, a wire can break under two equal and opposite pulling forces if the tension becomes too great. The breaking point of a wire depends on its material, thickness, and length. If the tension exceeds the wire's breaking point, it will snap and break.

4. How does the distance between the two forces affect the behavior of the wire?

The distance between the two forces, also known as the span, can greatly impact the behavior of the wire. A shorter span will result in a higher tension and more stretching of the wire, while a longer span will have less tension and less stretching. The span also affects the wire's stiffness and its ability to resist bending under the pulling forces.

5. What factors can influence the magnitude of the tension in a wire subjected to two equal and opposite pulling forces?

The magnitude of the tension in a wire subjected to two equal and opposite pulling forces can be influenced by several factors. These include the strength of the pulling forces, the type and thickness of the wire, the distance between the forces, and any external factors such as temperature or vibrations. Additionally, the wire's elasticity and ability to stretch and return to its original shape can also play a role in the tension magnitude.

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